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The angle will still be \(38^{\circ}\) regardless of magnification. 

I haven't done this  in  a  while, thebestchesscat, but I believe these are the answers

We have  three positive roots   (zeroes)....the sign on the y values changes three times

One between    0 and 1

One between     1 and 2

One between     2 and 3

The upper bound = 3   (I believe this  means that we have no other zeros  to  the  right of 3 )

We have one negative zero  at - 1

So...the degree ofthe poynomial is     1 negative root + 3 positive roots =   4th degree

Let P  be the other endpoint of the  diameter of the  circle

Angle SOT =  Angle SOP

Let the measure of SOP = x  = measure of  minor arc  SP    and      ROB = 2x = measure of minor arc  RB

We have 

Measure of  angle  RTB = (1/2) ( measure of  minor arc RB - measure of minor arc SP)

28  = (1/2) (2x - x)

28  = (1/2) x

x =  2 (28)  = 56  = measure of minor arc SP

And 2x = 112 = measure of minor arc RB

Measure  of minor arc RS =  180 - 112 - 56  =   12°

[ Certainly doesn't look like it from  the  given picture !!! ] 

Okay, thank you so much CPhill :))

(x -3)  ( [ x - (2+2i) ]  [ x - (2 - 2i)]   = 0

(x - 3)  [  x^2  - (2 - 2i)x - (2 + 2i)x + (2 + 2i) ( 2 -2i)  ]  =  0

(x - 3) [ x^2 - 4x + 4 - 4i^2 ]  =  0             [ i^2  = -1 ]

(x - 3) [ x^2 - 4x + 8]  =  0         The error occurs in the  second  step [ -4x instead of 4x ]

x^3 - 4x^2 + 8x  - 3x^3 + 12x - 24 = 0

x^3 - 7x^2 + 20x - 24 = 0

5n^2  + 19n + 16 >  0

This polynomial  is a parabola that turns upward and the  roots   are

[-19 + sqrt ( 19^2  - 4 * 5 * 16)] / 10   ≈ -1.26

And

[ -19 - sqrt (19^2 - 4 * 5 *16) ] / 10  ≈  -2.54

So.....The integer n =   -2  will not satisfy the inequality  [ at n =-2 the parabola lies  below the y axis ]

x^2 + 6x - 91  = 4x - 70

x^2 + 2x - 21  =  0

x^2 + 2x  =  21          complete the square on x

x^2 + 2x + 1  = 21 + 1

(x + 1)^2  =  22        take the  positive  root

x + 1  = sqrt (22)

x = sqrt (22)  - 1                (this is the only positive value of x that solves the equation)

Factor the polynomial in the  numerator  as  (c - 3) ( c + 9)

This gives us

(c+ 9)  + 2c =  23

3c + 9  = 23

3c =  14

c = 14 / 3

If x = 2 , then f(x)  =  0

If x = -1, then f(x)  =  1

The domain is  [ -1 , 2 ] 

See the following

Angle BCE  = 90    = Angle ADE

Because BC and DA are parallel, then angles EAD and EBC are opposite interior angles between these  parallels.....therefore they are equal

So by AA congruency, triangles EAD and EBC are similar

And  AD = 2 and BC = 1......therefore the ratio of the  sides = 2 : 1

Then BE =  [ 1 / (1 + 2) ] (AB)  =  ( 1/3)(5)  = 5/3

And by the Pythagoren Theorem, CE = sqrt ( BE^2 - BC^2) =  sqrt (5/3^2 - 1^2)  = sqrt (16/ 9) = 4/3

And AE =  5 - 5/3 =  10/3

And by the P Theorem again , ED = sqrt (  AE^2 - DA^2) =  sqrt [ (10/3)^2 - 2^2] =  sqrt [ 64/9] = 8/3

So CD =  CE + ED =  4/3 + 8/3 =  12/3 =  4

And ACBD is a parallelogram  with bases of 1,2   and  a height of 4

So [ ACBD ]  =  (1/2) (height) ( sum of the bases)  = (1/2) (4) ( 1 +2)  = 6 

i dont think thats correct cuz when you enter 4 as a value you get a complex number and you aren't supposed have them

Thanks!!!

Very nice, Bosco ....

Your reasoning  and   presentation  are  very good  !!!

I appreciate you so much :))

Equation of first circle

(x - 2)^2 + ( y + 1)^2 =  16   → x^2 - 4x + 4 + y^2 + 2y + 1  =  16

Equation of second  circle

(x + 2)^2  + (y - 5)^2 = 100 → x^2 + 4x + 4 + y^2 - 10y + 25 = 100

Subtract the second equation  from the  first

-8x + 12y -24 = -84

-8x + 12y  = - 60

-2x + 3y = -15

3y = 2x - 15

y=   [  (2/3)x - 5 ]     sub this into the first equation  for  y

(x - 2)^2  + [ (2/3)x - 5 + 1]^2  = 16

(x - 2)^2  + [ (2/3)x - 4]^2  = 16

x^2 - 4x + 4 + (4/9)x^2 - (16/3)x + 16 = 16

(13/9)x^2  -(28/3)x + 4  = 0

13x^2 - 84x + 36 = 0 

(13x -6) ( x - 6) = 0

Setting both factors to 0  and solving  for  x we have that

x = 6/13     and y = (2/3)(6/13) - 5  =  12/39 - 5  =  -61/13

x= 6    and y=  (2/3)(6) - 5 =  -1

(AB)^2 =  ( 6 - 6/13)^2 + ( -61/13 + 1)^2  =  (72/13)^2  + ( -48/13)^2  =  576 / 13

3x^2 + nx  + 72

n will be greatest when 72 is  factored as  72 * 1

Factor  as    (3x  +1) (x + 72)   →    n =  217 x  

x= 12  y = 6

xy = 72

The error occurs in the  second step

(x+ 4) ( 2x +1) (x+2)(x - 2) = 0

(2x^2 + 9x + 4) (x^2 - 4) = 0

2x^4 + 9x^3 + 4x^2 - 8x^2 - 36x -16 = 0

2x^4 + 9x^3 - 4x^2 -36x - 16  = 0

I misread the problem and made a mistake

I have created a diagram, and I added a few points for convenience. 

Notice that \(\overline{\rm BC} \perp \overleftrightarrow{\rm CD} \text{ and } \overline{\rm AD} \perp \overleftrightarrow{\rm CD}\) because of the Tangent Perpendicular to Radius Theroem. Since both segments are perpendicular to the same line, \(\overline{\rm AD} \parallel \overline{\rm BC}\). Quadrilateral ACBD has a pair of parallel sides and a pair of non-parallel sides, so this quadrilateral is classified as a trapezoid. The area formula for a trapezoid is \(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \) where b1 and b2 are the lengths of each parallel side and h is the perpendicular distance between b1 and b2 of the trapezoid. I have constructed the diagram such that \(\overline{\rm CE} \parallel \overline{\rm AB}\). Now, pay close attention to \(\triangle {\rm EDC}\). This is a right triangle, so we can use Pythagorean's Theorem to find \(\rm CD\), which is the height of the trapezoid.

\({\rm CD}^2 + {\rm ED}^2 = {\rm EC}^2 \\ {\rm CD}^2 + (2 + 1)^2 = 6^2 \\ {\rm CD}^2 = 36 - 9 \\ {\rm CD}^2 = 27 \\ {\rm CD} = 3\sqrt{3} \text{ or } {\rm CD} = -3\sqrt{3}\)

Since we are dealing with lengths, we should reject \({\rm CD} = -3\sqrt{3}\). Now, we have all the ingredients to find the area of the trapezoid ACBD.

\(A_{\text{trapezoid}} = \frac{1}{2}\left(b_1 + b_2\right)h \\ A_{\text{trapezoid}} = \frac{1}{2}(2 + 1)*3\sqrt{3} \\ A_{\text{trapezoid}} = \frac{9\sqrt{3}}{2} \approx 7.7942 \)